Lab 8: Population Genetics and Evolution

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1 Lab 8: Population Genetics and EvolutionThis may leave a bad taste in your mouth…

2 Pre-Lab OrientationRecall that the Hardy-Weinberg Equation helps us identify allele frequencies throughout a population.Given certain assumptions like “large population size,” “random mating,” et cetera…A great example for a classroom is ability to taste PTC (phenylthiocarbamide).This stuff is, oddly, used to grow transparent fish by inhibiting melanin production.

3 PTC Not everyone can taste PTC.In North America, 55% of people can taste it (it tastes quite bitter), while 45% cannot.Which one are you?Let’s take a taste, shall we? (you don’t have to)I’m going to wash my hands and give you a small piece.Place it on the tip of your tongue and wait a few seconds.If you are a PTC taster, you’ll get a bitter taste. If not, you’ll just taste something papery.Don’t swallow it. Just awkwardly peel it off your tongue and toss it in the trash.

4 Starting the lab…Take some time to read the Introduction (Page 80) and Exercise 8A while I’m coming around with PTC.Once we all taste the PTC strips, we’ll take some classroom data and record it in Table 8.1, then answer the Topics for Discussion on Page 82.Save Exercise 8B – Case Studies for another time.

5 Exercise 8B Now we start the Case Studies……so I’m sorry if this seems a little forward, but you need to go find a mate.A random mate – remember this is simulating Hardy-Weinberg equilibrium conditions. Find someone with whom you wouldn’t normally pair.Gender has no role here.Need a team of three? One of you is going to have to two-time it and act as partner to two others. Only write one result down, just like everyone else.

6 Case IStart by turning to the back page of your lab packet – the Data Page.Under Case I, note that we are simulating Initial Class Frequencies of:AA: 25%Aa: 50%aa: 25%Your Initial Genotype: AaOn your lab desks is a bag full of letters on card stock.These letters represent those same alleles that can be inherited – either A or a.

7 Case I Start by taking four total for each of you – two A and two a.These are the products of meiosis.Meiosis, if you forgot, is the production of gametes – haploid cells destined for reproduction.From one cell, four are produced, but they only have one set of chromosomes so there’s only one allele each.These are the only alleles you can pass on.

8 Case I Now put the cards into a pile, face down, and shuffle them.Take one card off the top. That’s one of the two alleles for your offspring. Have your partner do the same.These two alleles comprise the first offspring. One of you (and only one) should write the genotype in the data section on the last page (F1 Genotype).

9 Case INow put the letters back, repeat the process, develop a second offspring, and have the other partner write that down (F1 Genotype).Let’s record our data. (you don’t need to do this – just me)

10 Case I Okay…now the tricky part…Each of you needs to assume the role of the F1 generation.Leave the cards at the table but remember your genotype.Find a new mate and settle down at another lab table.Now that you have found a new mate, take a new set of cards.Remember that the letters represent the products of meiosis, so if you’re AA, take four A cards.If you’re Aa, take two A cards and two a cards.aa? Take four a cards.

11 Case I Class data time! What’s your genotype?Now you’re going to (randomly) find a new mate with the alleles you’ve determined.Find someone, then repeat the process.After each generation, pause so we can collect genotype data.Once we’re done with five generations, record the data in #4 after Case 1 on Page 93. Complete the questions and the Data Page section.

12 Case I DiscussionDid our allele frequencies change? Importantly, would we expect them to change?Is our population size large enough?Are we at Hardy-Weinberg equilibrium?

13 FYIWondering why our initial frequencies are 0.25 AA, 0.50 Aa, and 0.25 aa if everyone starts Aa?The answer is because you need to look forward a little.The Hardy-Weinberg equation tells us that, since p=0.5 and q=0.5 (we’re all heterozygous), p2 should be 0.25, 2pq is 0.50, and q2 is 0.25.Further, take a look at a Punnett Square for your first cross.

15 Case IINow we’re going to repeat Case I but change a condition – as in reality, not every genotype will have an equal chance of survival.An example is sickle-cell anemia, which can kill humans prior to reproduction if they are homozygous recessive.Each time you draw aa in this process, don’t record it; it doesn’t reproduce.You and the other parent must keep trying until you get a non-homozygous recessive offspring.Again, after five generations, we’re going to count frequencies, and then you’ll complete data and questions.

16 Case IIIBy now you know that sickle-cell anemia, while a disease, helps prevent a far worse disease in malaria.Individuals that are heterozygous for sickle-cell anemia have some sickle cells, but not enough to make them ill.At the same time, having those sickle-cells increases resistance to malaria because the parasites can’t infect the erythrocytes (red blood cells).

17 Case IIIWe’re going to simulate this heterozygote advantage with Case III.The procedure’s the same except:If your offspring is AA, flip a coin (or card).Heads = Does not survive (try again).Tails = Does survive.aa still doesn’t survive, by the way, and allele frequencies start the same (two A, two a, et cetera).We’ll do this for five generations, then record data, then do it for five more generations and record.And answer the questions as usual.

18 Case IV Case IV further explores the concept of genetic drift.Use the rules of Case I, but we’ll do so with a smaller population size.We will divide the class into three populations – no gene flow between groups!